Explanation

(a)
Let $y = f(\tan{x)\ and\ u = g(\sec{x)}}$
On differentiating w.r.t. $x,\$we get
$\frac{dy}{dx} = f^{'}(\tan{x)\sec^{2}x}$
and $\frac{du}{dx} = g^{'}(\sec{x)\sec{x\tan x}}$
$\therefore\ \frac{dy}{du} = \frac{dy\ /dx}{du\ /dx} = \frac{f^{'}(\tan{x)\sec^{2}x}}{g^{'}(\sec{x)\sec{x\ tan\ x}}}$
$\therefore\ \ \left( \frac{dy}{du} \right)_{x = \frac{\pi}{4}} = \frac{f^{'}\left( \tan\frac{\pi}{4} \right)}{g^{'}(\sec{\frac{\pi}{4})\sin\frac{\pi}{4}}}$
$= \frac{f^{'}(1).\sqrt{2}}{g^{'}(\sqrt{2)}} = \frac{2.\sqrt{2}}{4} = \frac{1}{\sqrt{2}}$